$\varepsilon$-product in Bierstedt's paper

Question

I am reading K.D.Bierstedt's paper Gewichtete Räume stetiger vektorwertiger Funktionen und das injektive Tensorprodukt. I. Journal für die reine und angewandte Mathematik 259 (1973): 186-210. It is written in German, and, perhaps, because of that I don't understand several details, and I hope that it will be proper to ask people here to clarify this to me.

As far as I understand, Bierstedt defines the $\varepsilon$-product $X\varepsilon Y$ (up to permutation $X$ and $Y$ which will be important for the comparison with Jarchow's definition below) of locally convex spaces $X$ and $Y$ as the space of linear continuous maps $$ \varphi:X_p'\to Y $$ where $X_p'$ is the space of linear continuous functionals $f:X\to\mathbb{C}$ equipped with the topology $p$ of uniform convergence on precompact sets $T\subseteq X$, by which I suppose he means totally bounded sets (i.e. such $T$ that for each neighbourhood of zero $U$ there is a finite set $A$ such that $T\subseteq U+A$).

And Bierstedt endows $X\varepsilon Y$ with the topology of uniform convergence on polars $U^\circ$ of neighbourhoods $U$ of zero in $X$.

And on page 197 he states the proposition ("Satz 9(3)") which I guess sounds in English like this:

If $X$ and $Y$ are complete and one of them has the approximation property, then $X\varepsilon Y$ coincides with the usual injective tensor product $X\check{\otimes}_\varepsilon Y$: $$X\varepsilon Y=X\check{\otimes}_\varepsilon Y$$

My first question is:

Q1. Do I understand this correctly?

I did not find this paper in English, and I don't speak German, that is why I have doubts.

If everything is correct, then a problem for me is that I don't understand how this is proved. Bierstedt gives a very meager explanation, which I don't understand, maybe because German is a problem for me. Everything would be more or less simple, but I stucked in the following

Lemma. If $X$ and $Y$ are complete, then $X\varepsilon Y$ is complete as well.

And my second question is

Q2. Is this lemma true? (And if yes, how is it proved?)

I have H.Jarchow's book, where he proves a similar fact (Theorem 16.1.5), but he defines $X\varepsilon Y$ differently, and what he proves seems to be not equivalent to Bierstedt's statament. In the case which is interesting for me, i.e. when $X$ and $Y$ are complete, the difference, as far as I understand, is that Jarchow endows $X'$ with another topology, which he denotes by $\gamma$, and this is the finest locally convex topology which coincides with the $X$-weak topology on polars $U^\circ$ of neighbourhoods of zero in $X$. And according to Jarchow, the space $X\varepsilon Y$ consists of other operators, the linear continuous mappings $$ \varphi:X_\gamma'\to Y. $$ So the impression is that Bierstedt's and Jarchow's $\varepsilon$-products are related to each other like this: $$ X\varepsilon_{\text{Bierstedt}} Y\subseteq X\varepsilon_{\text{Jarchow}} Y $$ I don't understand why there must be an equality here and why these two topologies $p$ and $\gamma$ on $X'$ must coincide.

Or, perhaps I miss something in this picture...

Can anybody explain to me how this problem is resolved? Why is Bierstedt's proposition true?

Answer 1

I don't have access to Bierstedt's article right now. Nevertheless, let me try to answer your question.

0. Yes, precompact sets are also called totally bounded.

1. The lemma is true and probably this is what Bierstedt says.

2. Yes, $X\varepsilon Y$ is complete whenever so are $X$ and $Y$. The proof should be quite standard (for a Cauchy net $T_i$ you get a pointwise limit because of the completeness of $Y$ and then you show that this limit is in the space and that the convergence is with respect to the topology of $X\varepsilon Y$).

3. This strange topology $\gamma$ on $X'$ (the finest locally convex topology which coincides with the weak$^*$-topology on all equicontinuous sets) has a another description (this is probably due to Grothendieck, at least, it is closely related to Grothendieck's construction of the completion) which you can see, e.g., in Köthe's Topolgical Vector Spaces I, §21.9(7) (in my edition, page 271): It is the topology of uniform convergence on all compact subsets of the completion of $X$ (which makes sense since the completion has the same dual). This explains the difference between Bierstedt's definition and the one in Jarchow: They coincide, if every compact subset of the completion is contained in the closure of a precompact subset of $X$. This is not always the case but, trivially, it holds for complete spaces which is, of course, the most important case.

I prefer Bierstedt's definition because the topology of precompact convergence seems more natural to me than $\gamma$.

Edit. Both definitions differ from the one of Laurent Schwartz (in his Théorie des Distribution à Valeurs Vectorielles) who endowed $X'$ with the topology of uniform convergence on absolutely convex compact sets -- Bierstedt denotes this original $\varepsilon$-product as $X\tilde\varepsilon Y$. Two advantages are the symmetry $X\tilde\varepsilon Y\cong Y\tilde\varepsilon X$ (the isomorphism is given by the transposition of operators) and $X\tilde\varepsilon \mathbb C\cong X$. Neither of the modifications of Bierstedt and Jarchow satisfies this because $X\varepsilon \mathbb C \cong \tilde X$ (the completion of $X$) but $\mathbb C\varepsilon X\cong X$. For complete spaces $X$ all three definitions coincide.